
#1
Feb1309, 12:51 AM

P: 391

Hi, everyone:
I am trying to see if it is true that if we are given f:C>C analytic ; C complex plane, then, to extend f to a function defined on C^ (Riemann Sphere) , i.e, to get: f^: C^ >C^ with f^_C =f i.e., the restriction of f^ to the complex plane agrees with f , If we need to define f(oo) =oo . I think the answer is yes. Here's what I have: We consider a 'hood ('hood:=neighborhood.) W of oo in C^ , which are complements of compact 'hoods K in C , together with {oo}, i.e., W=C\K U {oo} , for K compact in C. By Liouville's thm., f>oo on balls B(0;r) , as r>oo . And then by continuity, it would seem that we need f(oo)=oo, since W= CB(0;r) is a 'hood of oo. Alternatively, if we had an analytic map f on C , f would go to oo on balls B(0;r) as r>oo . Then if we used the stereo projection S: C>C^ , and push f forward by this projection, we would have Sof (composition)>oo . But this is still not rigorousenough. Any Ideas?. Thanks. P.S: If it bothers people to use regular ASCII, please let me know. I use ASCII as a way to force myself to keep things clear . But I can change if neccessary. 


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